A regular pentagon $ABCDE$ has diagonals $AC$ and $BE$ intersecting at $F$, and $AC$ and $BD$ intersecting at $G$. Show that \[\frac{AC}{AG} = \frac{AG}{AF} = \phi\] where $\phi$ is the ‘golden ratio’, $(1+\sqrt5)/2$. You may assume that $AB = AG$, as shown in Question 10.4.2. Show hint

$\Delta ABG \sim \Delta DAB$

There are a few choices of pairs of similar triangles in the diagram: $\Delta ABG \sim \Delta DAB \sim \Delta BFG$, and $\Delta GBC \sim \Delta BAC$ (not to mention all the other equivalent pairs like $\Delta FAB \sim \Delta ABE$). Any pair of them will lead you to the answer, but I’m just going to go over one of the choices.

All the diagonals of the pentagon are equal so $\Delta DAB$ is isosceles. It shares $\angle ABD$ with $\Delta ABG$, so the base angles of these two isosceles triangles are equal. Hence $\Delta ABG \sim \Delta DAB$. This gives \[\frac{AB}{BG} = \frac{DA}{AB}.\] Since the lengths we’re interested in are $AC$, $AG$, $AF$, and $FG$, it would help to write everything in terms of these lengths. $AB = AG$, $BG = AF$, and $DA = AC$, hence \[\frac{AG}{AF} = \frac{AC}{AG}.\] Let’s call this ratio $x$, then $AG = x AF$ and $AC = x AG$.

Now, how to find $x$? What other information can we get from the diagram? Note that $AC = AF + FC = AF + AG$. This will allow us to write everything in terms of one length, and hopefully solve for $x$. \begin{align*} AC &= AF + AG\\ x AG &= \frac{AG}{x} + AG\\ x &= \frac{1}{x} + 1\\ x^2 &= 1 + x\\ \end{align*} This quadratic equation gives $x = (1\pm\sqrt{5})/2$ as in Question 7.5.30. One of these solutions is negative and can be discarded, so $x = (1 + \sqrt{5})/2 = \phi$.